For one 120/100 Hz rectified (pulsating DC) peak, the conduction interval, with "reasonable" capacitances, is usually 1/4th to 1/5th of the total time, but can be much less with large capacitances, and can be 1/3rd of the total with less capacitance. At 1/3rd, the capacitance is often too small to prevent clipping at max rated power. For 60 Hz AC mains, with about 8.33 ms per 120 Hz peak, the charging pulses are typically about 2 ms long, for the minimum reservoir capacitances that allow the clipping threshold to be reached just as the rated peak output power is reached.

How can you count two peaks, when there is a conduction period for every peak (for a full-wave rectifier)?

The strategy, and precision required, is very different from that of audio circuits, but it gets results for supply impedance: 10 milliohms at 100Meg, 100 milliohms at 1GHz.

Interesting - I agree with the 'hierarchical capacitance' philosophy. Al electrolytic, tantalum, ceramic. Just I have an aversion to tants taking over the mid-frequency decoupling so instead I use outrageous quantities of ceramics and multiple paralleled lytics to bridge the gap...

The biggest problem with ceramics is how much capacitance they lose when fully biassed up - some are down to only 10% of the sticker value. Tricky to find reliable data on this and it varies between manufacturers quite a lot. TDK look to be one of the best.

__________________When irony first makes itself known in a young man's life, it can be like his first experience of getting drunk - Robertson Davies

For one 120/100 Hz rectified (pulsating DC) peak, the conduction interval, with "reasonable" capacitances, is usually 1/4th to 1/5th of the total time, but can be much less with large capacitances, and can be 1/3rd of the total with less capacitance. At 1/3rd, the capacitance is often too small to prevent clipping at max rated power. For 60 Hz AC mains, with about 8.33 ms per 120 Hz peak, the charging pulses are typically about 2 ms long, for the minimum reservoir capacitances that allow the clipping threshold to be reached just as the rated peak output power is reached.

How can you count two peaks, when there is a conduction period for every peak (for a full-wave rectifier)?

A 5% regulated transformer will reach the loaded peak voltage 18 degrees early due to float. It will begin to charge the capacitor at that time with the charging current increasing until the loaded peak current. It's post peak current then drops until the 5% float drops below the now higher voltage charged filter capacitor. Even an infinite value capacitor will not affect that.

Charging begins when the rising transformer output voltage minus the rectifier voltage(s) equals the falling capacitor voltage, which will be the minimum loaded voltage (not the peak loaded voltage).

Charging current will typically increase to way above the load's peak current, since the charging current has to supply as much charge, during the short charging interval, as the load uses during the entire 1/(2*fmains) period.

So how does the transformer regulation percentage affect the charging pulse length (or, rather, by how much does it affect it, and how can that be calculated)?

So if I understand what is being said here we have a pulsed current flowing to charge the capacitor as the load draws down the capacitors charge. Now is there any way to increase the charging time or is that impossible, because we are trying to keep up with the current demands of the output devices. How would a smps play into this conductance angle, or would that still have the identical pulse charging problems. The last question is is it really a problem just the way it is?

Well for one the first issue was my bad... Long day I didn't rennet correctly my first post.

If you take any power transformer and measure the output voltage without any load it will be greater than when it is loaded to the rated current. A small cheap transformer may go as high as 20%. A well built power transformer suitable for an audio power amplifier is almost always rated at 5%. This is due to the resistance of the windings and other losses that occur under load.

So until the filter capacitor begins to draw current the output voltage will behave as if unloaded and be 5% higher. Now the rectifier also has a lower voltage drop at lower currents but as the voltage drop rise much more quickly with current it really can be considered a constant drop at all times so it doesn't influence the charging time available.

So the result is that current begins to flow as soon as the output voltage goes above the stored voltage. However as the charging current increases so does the power loss from the internal loses. At the peak voltage the transformer is fully loaded. Due to the peak current being higher than the rated current the actual peak voltage is lower than the rated voltage!

After the peak voltage the current begins to drop and the transformer losses decrease so even the post peak voltage begins to rise.

The biggest problem with ceramics is how much capacitance they lose when fully biassed up - some are down to only 10% of the sticker value. Tricky to find reliable data on this and it varies between manufacturers quite a lot. TDK look to be one of the best.

But, the actual capacitance is almost irrelevant ... at the high frequencies you're using the overlapping of the impedance minimums at resonance to get the job done - sufficient of around the right value should do it.

Gootee,
If the charging pulse stops 5 degrees after the halfwave peak, then a 60degree conduction angle requires conduction to begin 55degrees before the peak of that halfwave.

At 55 degrees before the peak, the sinewave has reaching about 57% of the transformer voltage.
If conduction begins at that 57% value then the capacitor voltage must have dropped to the 57% at the end of the previous non-conduction period.

If I had a ripple on my smoothing capacitor that dropped to 57% of the transformer voltage, then I have failed to design and build an adequate PSU.

The conduction angle cannot be 60degrees on any properly designed and built amplifier PSU. Ripple that bad is never tolerated.

Could someone estimate the Amperes per mF of smoothiong that would ensue to give a 60degree conduction angle. I'll guess it is no where near the typical 1A/mF to ½A/mF that we often use as a guide for smoothing capacitance.

__________________
regards Andrew T.
Sent from my desktop computer using a keyboard

It is 60 degrees of the total 360. But at maximum current it is a bit longer. To get it shorter you must have a transformer with more capacity than the load requires.

The float doesn't just give you extra charge time on the front end but also on the discharge side.

If you have 5% ripple then you start charging at .9 or 26 degrees before the peak at 5% regulation. After the peak the slope of the sine wave is flatter than the discharge slope for a bit less than 3 degrees then you follow the float for another 18 degrees. That would be about the longest charge time.

Now if your transformer is rated for ten times the capacity required your output voltage would be 5% (almost) higher and your float would be .5%. That is only .3 degrees on the front end for an infinite capacitor. At 5% ripple you would add 18 degrees to that on the front and 3 degrees on the back so the total would be 22 degrees of charging per half cycle.

The charging time is greatly dependent on transformer regulation. It is only with a lot of ripple that the capacitance becomes dominant.